Source code for sympy.solvers.diophantine

from __future__ import print_function, division

from sympy.core.add import Add
from sympy.core.compatibility import as_int, is_sequence, range
from sympy.core.exprtools import factor_terms
from sympy.core.function import _mexpand
from sympy.core.mul import Mul
from sympy.core.numbers import Rational
from sympy.core.numbers import igcdex, ilcm, igcd
from sympy.core.power import integer_nthroot, isqrt
from sympy.core.relational import Eq
from sympy.core.singleton import S
from sympy.core.symbol import Symbol, symbols
from sympy.functions.elementary.complexes import sign
from sympy.functions.elementary.integers import floor
from sympy.functions.elementary.miscellaneous import sqrt
from sympy.matrices.dense import MutableDenseMatrix as Matrix
from sympy.ntheory.factor_ import (
    divisors, factorint, multiplicity, perfect_power)
from sympy.ntheory.generate import nextprime
from sympy.ntheory.primetest import is_square, isprime
from sympy.ntheory.residue_ntheory import sqrt_mod
from sympy.polys.polyerrors import GeneratorsNeeded
from sympy.polys.polytools import Poly, factor_list
from sympy.simplify.simplify import signsimp
from sympy.solvers.solvers import check_assumptions
from sympy.solvers.solveset import solveset_real
from sympy.utilities import default_sort_key, numbered_symbols
from sympy.utilities.misc import filldedent



# these are imported with 'from sympy.solvers.diophantine import *
__all__ = ['diophantine', 'classify_diop']


# these types are known (but not necessarily handled)
diop_known = {
    "binary_quadratic",
    "cubic_thue",
    "general_pythagorean",
    "general_sum_of_even_powers",
    "general_sum_of_squares",
    "homogeneous_general_quadratic",
    "homogeneous_ternary_quadratic",
    "homogeneous_ternary_quadratic_normal",
    "inhomogeneous_general_quadratic",
    "inhomogeneous_ternary_quadratic",
    "linear",
    "univariate"}


def _is_int(i):
    try:
        as_int(i)
        return True
    except ValueError:
        pass


def _sorted_tuple(*i):
    return tuple(sorted(i))


def _remove_gcd(*x):
    try:
        g = igcd(*x)
        return tuple([i//g for i in x])
    except ValueError:
        return x
    except TypeError:
        raise TypeError('_remove_gcd(a,b,c) or _remove_gcd(*container)')


def _rational_pq(a, b):
    # return `(numer, denom)` for a/b; sign in numer and gcd removed
    return _remove_gcd(sign(b)*a, abs(b))


def _nint_or_floor(p, q):
    # return nearest int to p/q; in case of tie return floor(p/q)
    w, r = divmod(p, q)
    if abs(r) <= abs(q)//2:
        return w
    return w + 1


def _odd(i):
    return i % 2 != 0


def _even(i):
    return i % 2 == 0


[docs]def diophantine(eq, param=symbols("t", integer=True), syms=None, permute=False): """ Simplify the solution procedure of diophantine equation ``eq`` by converting it into a product of terms which should equal zero. For example, when solving, `x^2 - y^2 = 0` this is treated as `(x + y)(x - y) = 0` and `x + y = 0` and `x - y = 0` are solved independently and combined. Each term is solved by calling ``diop_solve()``. Output of ``diophantine()`` is a set of tuples. The elements of the tuple are the solutions for each variable in the equation and are arranged according to the alphabetic ordering of the variables. e.g. For an equation with two variables, `a` and `b`, the first element of the tuple is the solution for `a` and the second for `b`. Usage ===== ``diophantine(eq, t, syms)``: Solve the diophantine equation ``eq``. ``t`` is the optional parameter to be used by ``diop_solve()``. ``syms`` is an optional list of symbols which determines the order of the elements in the returned tuple. By default, only the base solution is returned. If ``permute`` is set to True then permutations of the base solution and/or permutations of the signs of the values will be returned when applicable. >>> from sympy.solvers.diophantine import diophantine >>> from sympy.abc import a, b >>> eq = a**4 + b**4 - (2**4 + 3**4) >>> diophantine(eq) {(2, 3)} >>> diophantine(eq, permute=True) {(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)} Details ======= ``eq`` should be an expression which is assumed to be zero. ``t`` is the parameter to be used in the solution. Examples ======== >>> from sympy.abc import x, y, z >>> diophantine(x**2 - y**2) {(t_0, -t_0), (t_0, t_0)} >>> diophantine(x*(2*x + 3*y - z)) {(0, n1, n2), (t_0, t_1, 2*t_0 + 3*t_1)} >>> diophantine(x**2 + 3*x*y + 4*x) {(0, n1), (3*t_0 - 4, -t_0)} See Also ======== diop_solve() sympy.utilities.iterables.permute_signs sympy.utilities.iterables.signed_permutations """ from sympy.utilities.iterables import ( subsets, permute_signs, signed_permutations) if isinstance(eq, Eq): eq = eq.lhs - eq.rhs try: var = list(eq.expand(force=True).free_symbols) var.sort(key=default_sort_key) if syms: if not is_sequence(syms): raise TypeError( 'syms should be given as a sequence, e.g. a list') syms = [i for i in syms if i in var] if syms != var: dict_sym_index = dict(zip(syms, range(len(syms)))) return {tuple([t[dict_sym_index[i]] for i in var]) for t in diophantine(eq, param)} n, d = eq.as_numer_denom() if not n.free_symbols: return set() if d.free_symbols: dsol = diophantine(d) good = diophantine(n) - dsol return {s for s in good if _mexpand(d.subs(zip(var, s)))} else: eq = n eq = factor_terms(eq) assert not eq.is_number eq = eq.as_independent(*var, as_Add=False)[1] p = Poly(eq) assert not any(g.is_number for g in p.gens) eq = p.as_expr() assert eq.is_polynomial() except (GeneratorsNeeded, AssertionError, AttributeError): raise TypeError(filldedent(''' Equation should be a polynomial with Rational coefficients.''')) # permute only sign do_permute_signs = False # permute sign and values do_permute_signs_var = False # permute few signs permute_few_signs = False try: # if we know that factoring should not be attempted, skip # the factoring step v, c, t = classify_diop(eq) # check for permute sign if permute: len_var = len(v) permute_signs_for = [ 'general_sum_of_squares', 'general_sum_of_even_powers'] permute_signs_check = [ 'homogeneous_ternary_quadratic', 'homogeneous_ternary_quadratic_normal', 'binary_quadratic'] if t in permute_signs_for: do_permute_signs_var = True elif t in permute_signs_check: # if all the variables in eq have even powers # then do_permute_sign = True if len_var == 3: var_mul = list(subsets(v, 2)) # here var_mul is like [(x, y), (x, z), (y, z)] xy_coeff = True x_coeff = True var1_mul_var2 = map(lambda a: a[0]*a[1], var_mul) # if coeff(y*z), coeff(y*x), coeff(x*z) is not 0 then # `xy_coeff` => True and do_permute_sign => False. # Means no permuted solution. for v1_mul_v2 in var1_mul_var2: try: coeff = c[v1_mul_v2] except KeyError: coeff = 0 xy_coeff = bool(xy_coeff) and bool(coeff) var_mul = list(subsets(v, 1)) # here var_mul is like [(x,), (y, )] for v1 in var_mul: try: coeff = c[var[0]] except KeyError: coeff = 0 x_coeff = bool(x_coeff) and bool(coeff) if not any([xy_coeff, x_coeff]): # means only x**2, y**2, z**2, const is present do_permute_signs = True elif not x_coeff: permute_few_signs = True elif len_var == 2: var_mul = list(subsets(v, 2)) # here var_mul is like [(x, y)] xy_coeff = True x_coeff = True var1_mul_var2 = map(lambda x: x[0]*x[1], var_mul) for v1_mul_v2 in var1_mul_var2: try: coeff = c[v1_mul_v2] except KeyError: coeff = 0 xy_coeff = bool(xy_coeff) and bool(coeff) var_mul = list(subsets(v, 1)) # here var_mul is like [(x,), (y, )] for v1 in var_mul: try: coeff = c[var[0]] except KeyError: coeff = 0 x_coeff = bool(x_coeff) and bool(coeff) if not any([xy_coeff, x_coeff]): # means only x**2, y**2 and const is present # so we can get more soln by permuting this soln. do_permute_signs = True elif not x_coeff: # when coeff(x), coeff(y) is not present then signs of # x, y can be permuted such that their sign are same # as sign of x*y. # e.g 1. (x_val,y_val)=> (x_val,y_val), (-x_val,-y_val) # 2. (-x_vall, y_val)=> (-x_val,y_val), (x_val,-y_val) permute_few_signs = True if t == 'general_sum_of_squares': # trying to factor such expressions will sometimes hang terms = [(eq, 1)] else: raise TypeError except (TypeError, NotImplementedError): terms = factor_list(eq)[1] sols = set([]) for term in terms: base, _ = term var_t, _, eq_type = classify_diop(base, _dict=False) _, base = signsimp(base, evaluate=False).as_coeff_Mul() solution = diop_solve(base, param) if eq_type in [ "linear", "homogeneous_ternary_quadratic", "homogeneous_ternary_quadratic_normal", "general_pythagorean"]: sols.add(merge_solution(var, var_t, solution)) elif eq_type in [ "binary_quadratic", "general_sum_of_squares", "general_sum_of_even_powers", "univariate"]: for sol in solution: sols.add(merge_solution(var, var_t, sol)) else: raise NotImplementedError('unhandled type: %s' % eq_type) # remove null merge results if () in sols: sols.remove(()) null = tuple([0]*len(var)) # if there is no solution, return trivial solution if not sols and eq.subs(zip(var, null)) is S.Zero: sols.add(null) final_soln = set([]) for sol in sols: if all(_is_int(s) for s in sol): if do_permute_signs: permuted_sign = set(permute_signs(sol)) final_soln.update(permuted_sign) elif permute_few_signs: lst = list(permute_signs(sol)) lst = list(filter(lambda x: x[0]*x[1] == sol[1]*sol[0], lst)) permuted_sign = set(lst) final_soln.update(permuted_sign) elif do_permute_signs_var: permuted_sign_var = set(signed_permutations(sol)) final_soln.update(permuted_sign_var) else: final_soln.add(sol) else: final_soln.add(sol) return final_soln
[docs]def merge_solution(var, var_t, solution): """ This is used to construct the full solution from the solutions of sub equations. For example when solving the equation `(x - y)(x^2 + y^2 - z^2) = 0`, solutions for each of the equations `x - y = 0` and `x^2 + y^2 - z^2` are found independently. Solutions for `x - y = 0` are `(x, y) = (t, t)`. But we should introduce a value for z when we output the solution for the original equation. This function converts `(t, t)` into `(t, t, n_{1})` where `n_{1}` is an integer parameter. """ sol = [] if None in solution: return () solution = iter(solution) params = numbered_symbols("n", integer=True, start=1) for v in var: if v in var_t: sol.append(next(solution)) else: sol.append(next(params)) for val, symb in zip(sol, var): if check_assumptions(val, **symb.assumptions0) is False: return tuple() return tuple(sol)
[docs]def diop_solve(eq, param=symbols("t", integer=True)): """ Solves the diophantine equation ``eq``. Unlike ``diophantine()``, factoring of ``eq`` is not attempted. Uses ``classify_diop()`` to determine the type of the equation and calls the appropriate solver function. Usage ===== ``diop_solve(eq, t)``: Solve diophantine equation, ``eq`` using ``t`` as a parameter if needed. Details ======= ``eq`` should be an expression which is assumed to be zero. ``t`` is a parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine import diop_solve >>> from sympy.abc import x, y, z, w >>> diop_solve(2*x + 3*y - 5) (3*t_0 - 5, -2*t_0 + 5) >>> diop_solve(4*x + 3*y - 4*z + 5) (t_0, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5) >>> diop_solve(x + 3*y - 4*z + w - 6) (t_0, t_0 + t_1, 6*t_0 + 5*t_1 + 4*t_2 - 6, 5*t_0 + 4*t_1 + 3*t_2 - 6) >>> diop_solve(x**2 + y**2 - 5) {(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)} See Also ======== diophantine() """ var, coeff, eq_type = classify_diop(eq, _dict=False) if eq_type == "linear": return _diop_linear(var, coeff, param) elif eq_type == "binary_quadratic": return _diop_quadratic(var, coeff, param) elif eq_type == "homogeneous_ternary_quadratic": x_0, y_0, z_0 = _diop_ternary_quadratic(var, coeff) return _parametrize_ternary_quadratic( (x_0, y_0, z_0), var, coeff) elif eq_type == "homogeneous_ternary_quadratic_normal": x_0, y_0, z_0 = _diop_ternary_quadratic_normal(var, coeff) return _parametrize_ternary_quadratic( (x_0, y_0, z_0), var, coeff) elif eq_type == "general_pythagorean": return _diop_general_pythagorean(var, coeff, param) elif eq_type == "univariate": return set([(int(i),) for i in solveset_real( eq, var[0]).intersect(S.Integers)]) elif eq_type == "general_sum_of_squares": return _diop_general_sum_of_squares(var, -int(coeff[1]), limit=S.Infinity) elif eq_type == "general_sum_of_even_powers": for k in coeff.keys(): if k.is_Pow and coeff[k]: p = k.exp return _diop_general_sum_of_even_powers(var, p, -int(coeff[1]), limit=S.Infinity) if eq_type is not None and eq_type not in diop_known: raise ValueError(filldedent(''' Alhough this type of equation was identified, it is not yet handled. It should, however, be listed in `diop_known` at the top of this file. Developers should see comments at the end of `classify_diop`. ''')) # pragma: no cover else: raise NotImplementedError( 'No solver has been written for %s.' % eq_type)
[docs]def classify_diop(eq, _dict=True): # docstring supplied externally try: var = list(eq.free_symbols) assert var except (AttributeError, AssertionError): raise ValueError('equation should have 1 or more free symbols') var.sort(key=default_sort_key) eq = eq.expand(force=True) coeff = eq.as_coefficients_dict() if not all(_is_int(c) for c in coeff.values()): raise TypeError("Coefficients should be Integers") diop_type = None total_degree = Poly(eq).total_degree() homogeneous = 1 not in coeff if total_degree == 1: diop_type = "linear" elif len(var) == 1: diop_type = "univariate" elif total_degree == 2 and len(var) == 2: diop_type = "binary_quadratic" elif total_degree == 2 and len(var) == 3 and homogeneous: if set(coeff) & set(var): diop_type = "inhomogeneous_ternary_quadratic" else: nonzero = [k for k in coeff if coeff[k]] if len(nonzero) == 3 and all(i**2 in nonzero for i in var): diop_type = "homogeneous_ternary_quadratic_normal" else: diop_type = "homogeneous_ternary_quadratic" elif total_degree == 2 and len(var) >= 3: if set(coeff) & set(var): diop_type = "inhomogeneous_general_quadratic" else: # there may be Pow keys like x**2 or Mul keys like x*y if any(k.is_Mul for k in coeff): # cross terms if not homogeneous: diop_type = "inhomogeneous_general_quadratic" else: diop_type = "homogeneous_general_quadratic" else: # all squares: x**2 + y**2 + ... + constant if all(coeff[k] == 1 for k in coeff if k != 1): diop_type = "general_sum_of_squares" elif all(is_square(abs(coeff[k])) for k in coeff): if abs(sum(sign(coeff[k]) for k in coeff)) == \ len(var) - 2: # all but one has the same sign # e.g. 4*x**2 + y**2 - 4*z**2 diop_type = "general_pythagorean" elif total_degree == 3 and len(var) == 2: diop_type = "cubic_thue" elif (total_degree > 3 and total_degree % 2 == 0 and all(k.is_Pow and k.exp == total_degree for k in coeff if k != 1)): if all(coeff[k] == 1 for k in coeff if k != 1): diop_type = 'general_sum_of_even_powers' if diop_type is not None: return var, dict(coeff) if _dict else coeff, diop_type # new diop type instructions # -------------------------- # if this error raises and the equation *can* be classified, # * it should be identified in the if-block above # * the type should be added to the diop_known # if a solver can be written for it, # * a dedicated handler should be written (e.g. diop_linear) # * it should be passed to that handler in diop_solve raise NotImplementedError(filldedent(''' This equation is not yet recognized or else has not been simplified sufficiently to put it in a form recognized by diop_classify().'''))
classify_diop.func_doc = ''' Helper routine used by diop_solve() to find information about ``eq``. Returns a tuple containing the type of the diophantine equation along with the variables (free symbols) and their coefficients. Variables are returned as a list and coefficients are returned as a dict with the key being the respective term and the constant term is keyed to 1. The type is one of the following: * %s Usage ===== ``classify_diop(eq)``: Return variables, coefficients and type of the ``eq``. Details ======= ``eq`` should be an expression which is assumed to be zero. ``_dict`` is for internal use: when True (default) a dict is returned, otherwise a defaultdict which supplies 0 for missing keys is returned. Examples ======== >>> from sympy.solvers.diophantine import classify_diop >>> from sympy.abc import x, y, z, w, t >>> classify_diop(4*x + 6*y - 4) ([x, y], {1: -4, x: 4, y: 6}, 'linear') >>> classify_diop(x + 3*y -4*z + 5) ([x, y, z], {1: 5, x: 1, y: 3, z: -4}, 'linear') >>> classify_diop(x**2 + y**2 - x*y + x + 5) ([x, y], {1: 5, x: 1, x**2: 1, y**2: 1, x*y: -1}, 'binary_quadratic') ''' % ('\n * '.join(sorted(diop_known)))
[docs]def diop_linear(eq, param=symbols("t", integer=True)): """ Solves linear diophantine equations. A linear diophantine equation is an equation of the form `a_{1}x_{1} + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. Usage ===== ``diop_linear(eq)``: Returns a tuple containing solutions to the diophantine equation ``eq``. Values in the tuple is arranged in the same order as the sorted variables. Details ======= ``eq`` is a linear diophantine equation which is assumed to be zero. ``param`` is the parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine import diop_linear >>> from sympy.abc import x, y, z, t >>> diop_linear(2*x - 3*y - 5) # solves equation 2*x - 3*y - 5 == 0 (3*t_0 - 5, 2*t_0 - 5) Here x = -3*t_0 - 5 and y = -2*t_0 - 5 >>> diop_linear(2*x - 3*y - 4*z -3) (t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3) See Also ======== diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean(), diop_general_sum_of_squares() """ from sympy.core.function import count_ops var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "linear": return _diop_linear(var, coeff, param)
def _diop_linear(var, coeff, param): """ Solves diophantine equations of the form: a_0*x_0 + a_1*x_1 + ... + a_n*x_n == c Note that no solution exists if gcd(a_0, ..., a_n) doesn't divide c. """ if 1 in coeff: # negate coeff[] because input is of the form: ax + by + c == 0 # but is used as: ax + by == -c c = -coeff[1] else: c = 0 # Some solutions will have multiple free variables in their solutions. if param is None: params = [symbols('t')]*len(var) else: temp = str(param) + "_%i" params = [symbols(temp % i, integer=True) for i in range(len(var))] if len(var) == 1: q, r = divmod(c, coeff[var[0]]) if not r: return (q,) else: return (None,) ''' base_solution_linear() can solve diophantine equations of the form: a*x + b*y == c We break down multivariate linear diophantine equations into a series of bivariate linear diophantine equations which can then be solved individually by base_solution_linear(). Consider the following: a_0*x_0 + a_1*x_1 + a_2*x_2 == c which can be re-written as: a_0*x_0 + g_0*y_0 == c where g_0 == gcd(a_1, a_2) and y == (a_1*x_1)/g_0 + (a_2*x_2)/g_0 This leaves us with two binary linear diophantine equations. For the first equation: a == a_0 b == g_0 c == c For the second: a == a_1/g_0 b == a_2/g_0 c == the solution we find for y_0 in the first equation. The arrays A and B are the arrays of integers used for 'a' and 'b' in each of the n-1 bivariate equations we solve. ''' A = [coeff[v] for v in var] B = [] if len(var) > 2: B.append(igcd(A[-2], A[-1])) A[-2] = A[-2] // B[0] A[-1] = A[-1] // B[0] for i in range(len(A) - 3, 0, -1): gcd = igcd(B[0], A[i]) B[0] = B[0] // gcd A[i] = A[i] // gcd B.insert(0, gcd) B.append(A[-1]) ''' Consider the trivariate linear equation: 4*x_0 + 6*x_1 + 3*x_2 == 2 This can be re-written as: 4*x_0 + 3*y_0 == 2 where y_0 == 2*x_1 + x_2 (Note that gcd(3, 6) == 3) The complete integral solution to this equation is: x_0 == 2 + 3*t_0 y_0 == -2 - 4*t_0 where 't_0' is any integer. Now that we have a solution for 'x_0', find 'x_1' and 'x_2': 2*x_1 + x_2 == -2 - 4*t_0 We can then solve for '-2' and '-4' independently, and combine the results: 2*x_1a + x_2a == -2 x_1a == 0 + t_0 x_2a == -2 - 2*t_0 2*x_1b + x_2b == -4*t_0 x_1b == 0*t_0 + t_1 x_2b == -4*t_0 - 2*t_1 ==> x_1 == t_0 + t_1 x_2 == -2 - 6*t_0 - 2*t_1 where 't_0' and 't_1' are any integers. Note that: 4*(2 + 3*t_0) + 6*(t_0 + t_1) + 3*(-2 - 6*t_0 - 2*t_1) == 2 for any integral values of 't_0', 't_1'; as required. This method is generalised for many variables, below. ''' solutions = [] for i in range(len(B)): tot_x, tot_y = [], [] for j, arg in enumerate(Add.make_args(c)): if arg.is_Integer: # example: 5 -> k = 5 k, p = arg, S.One pnew = params[0] else: # arg is a Mul or Symbol # example: 3*t_1 -> k = 3 # example: t_0 -> k = 1 k, p = arg.as_coeff_Mul() pnew = params[params.index(p) + 1] sol = sol_x, sol_y = base_solution_linear(k, A[i], B[i], pnew) if p is S.One: if None in sol: return tuple([None]*len(var)) else: # convert a + b*pnew -> a*p + b*pnew if isinstance(sol_x, Add): sol_x = sol_x.args[0]*p + sol_x.args[1] if isinstance(sol_y, Add): sol_y = sol_y.args[0]*p + sol_y.args[1] tot_x.append(sol_x) tot_y.append(sol_y) solutions.append(Add(*tot_x)) c = Add(*tot_y) solutions.append(c) if param is None: # just keep the additive constant (i.e. replace t with 0) solutions = [i.as_coeff_Add()[0] for i in solutions] return tuple(solutions)
[docs]def base_solution_linear(c, a, b, t=None): """ Return the base solution for the linear equation, `ax + by = c`. Used by ``diop_linear()`` to find the base solution of a linear Diophantine equation. If ``t`` is given then the parametrized solution is returned. Usage ===== ``base_solution_linear(c, a, b, t)``: ``a``, ``b``, ``c`` are coefficients in `ax + by = c` and ``t`` is the parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine import base_solution_linear >>> from sympy.abc import t >>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5 (-5, 5) >>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0 (0, 0) >>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5 (3*t - 5, -2*t + 5) >>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0 (7*t, -5*t) """ a, b, c = _remove_gcd(a, b, c) if c == 0: if t is not None: if b < 0: t = -t return (b*t , -a*t) else: return (0, 0) else: x0, y0, d = igcdex(abs(a), abs(b)) x0 *= sign(a) y0 *= sign(b) if divisible(c, d): if t is not None: if b < 0: t = -t return (c*x0 + b*t, c*y0 - a*t) else: return (c*x0, c*y0) else: return (None, None)
[docs]def divisible(a, b): """ Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise. """ return not a % b
[docs]def diop_quadratic(eq, param=symbols("t", integer=True)): """ Solves quadratic diophantine equations. i.e. equations of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`. Returns a set containing the tuples `(x, y)` which contains the solutions. If there are no solutions then `(None, None)` is returned. Usage ===== ``diop_quadratic(eq, param)``: ``eq`` is a quadratic binary diophantine equation. ``param`` is used to indicate the parameter to be used in the solution. Details ======= ``eq`` should be an expression which is assumed to be zero. ``param`` is a parameter to be used in the solution. Examples ======== >>> from sympy.abc import x, y, t >>> from sympy.solvers.diophantine import diop_quadratic >>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t) {(-1, -1)} References ========== .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], Available: http://www.alpertron.com.ar/METHODS.HTM .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: http://www.jpr2718.org/ax2p.pdf See Also ======== diop_linear(), diop_ternary_quadratic(), diop_general_sum_of_squares(), diop_general_pythagorean() """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "binary_quadratic": return _diop_quadratic(var, coeff, param)
def _diop_quadratic(var, coeff, t): x, y = var A = coeff[x**2] B = coeff[x*y] C = coeff[y**2] D = coeff[x] E = coeff[y] F = coeff[1] A, B, C, D, E, F = [as_int(i) for i in _remove_gcd(A, B, C, D, E, F)] # (1) Simple-Hyperbolic case: A = C = 0, B != 0 # In this case equation can be converted to (Bx + E)(By + D) = DE - BF # We consider two cases; DE - BF = 0 and DE - BF != 0 # More details, http://www.alpertron.com.ar/METHODS.HTM#SHyperb sol = set([]) discr = B**2 - 4*A*C if A == 0 and C == 0 and B != 0: if D*E - B*F == 0: q, r = divmod(E, B) if not r: sol.add((-q, t)) q, r = divmod(D, B) if not r: sol.add((t, -q)) else: div = divisors(D*E - B*F) div = div + [-term for term in div] for d in div: x0, r = divmod(d - E, B) if not r: q, r = divmod(D*E - B*F, d) if not r: y0, r = divmod(q - D, B) if not r: sol.add((x0, y0)) # (2) Parabolic case: B**2 - 4*A*C = 0 # There are two subcases to be considered in this case. # sqrt(c)D - sqrt(a)E = 0 and sqrt(c)D - sqrt(a)E != 0 # More Details, http://www.alpertron.com.ar/METHODS.HTM#Parabol elif discr == 0: if A == 0: s = _diop_quadratic([y, x], coeff, t) for soln in s: sol.add((soln[1], soln[0])) else: g = sign(A)*igcd(A, C) a = A // g b = B // g c = C // g e = sign(B/A) sqa = isqrt(a) sqc = isqrt(c) _c = e*sqc*D - sqa*E if not _c: z = symbols("z", real=True) eq = sqa*g*z**2 + D*z + sqa*F roots = solveset_real(eq, z).intersect(S.Integers) for root in roots: ans = diop_solve(sqa*x + e*sqc*y - root) sol.add((ans[0], ans[1])) elif _is_int(c): solve_x = lambda u: -e*sqc*g*_c*t**2 - (E + 2*e*sqc*g*u)*t\ - (e*sqc*g*u**2 + E*u + e*sqc*F) // _c solve_y = lambda u: sqa*g*_c*t**2 + (D + 2*sqa*g*u)*t \ + (sqa*g*u**2 + D*u + sqa*F) // _c for z0 in range(0, abs(_c)): # Check if the coefficients of y and x obtained are integers or not if (divisible(sqa*g*z0**2 + D*z0 + sqa*F, _c) and divisible(e*sqc**g*z0**2 + E*z0 + e*sqc*F, _c)): sol.add((solve_x(z0), solve_y(z0))) # (3) Method used when B**2 - 4*A*C is a square, is described in p. 6 of the below paper # by John P. Robertson. # http://www.jpr2718.org/ax2p.pdf elif is_square(discr): if A != 0: r = sqrt(discr) u, v = symbols("u, v", integer=True) eq = _mexpand( 4*A*r*u*v + 4*A*D*(B*v + r*u + r*v - B*u) + 2*A*4*A*E*(u - v) + 4*A*r*4*A*F) solution = diop_solve(eq, t) for s0, t0 in solution: num = B*t0 + r*s0 + r*t0 - B*s0 x_0 = S(num)/(4*A*r) y_0 = S(s0 - t0)/(2*r) if isinstance(s0, Symbol) or isinstance(t0, Symbol): if check_param(x_0, y_0, 4*A*r, t) != (None, None): ans = check_param(x_0, y_0, 4*A*r, t) sol.add((ans[0], ans[1])) elif x_0.is_Integer and y_0.is_Integer: if is_solution_quad(var, coeff, x_0, y_0): sol.add((x_0, y_0)) else: s = _diop_quadratic(var[::-1], coeff, t) # Interchange x and y while s: # | sol.add(s.pop()[::-1]) # and solution <--------+ # (4) B**2 - 4*A*C > 0 and B**2 - 4*A*C not a square or B**2 - 4*A*C < 0 else: P, Q = _transformation_to_DN(var, coeff) D, N = _find_DN(var, coeff) solns_pell = diop_DN(D, N) if D < 0: for x0, y0 in solns_pell: for x in [-x0, x0]: for y in [-y0, y0]: s = P*Matrix([x, y]) + Q try: sol.add(tuple([as_int(_) for _ in s])) except ValueError: pass else: # In this case equation can be transformed into a Pell equation solns_pell = set(solns_pell) for X, Y in list(solns_pell): solns_pell.add((-X, -Y)) a = diop_DN(D, 1) T = a[0][0] U = a[0][1] if all(_is_int(_) for _ in P[:4] + Q[:2]): for r, s in solns_pell: _a = (r + s*sqrt(D))*(T + U*sqrt(D))**t _b = (r - s*sqrt(D))*(T - U*sqrt(D))**t x_n = _mexpand(S(_a + _b)/2) y_n = _mexpand(S(_a - _b)/(2*sqrt(D))) s = P*Matrix([x_n, y_n]) + Q sol.add(tuple(s)) else: L = ilcm(*[_.q for _ in P[:4] + Q[:2]]) k = 1 T_k = T U_k = U while (T_k - 1) % L != 0 or U_k % L != 0: T_k, U_k = T_k*T + D*U_k*U, T_k*U + U_k*T k += 1 for X, Y in solns_pell: for i in range(k): if all(_is_int(_) for _ in P*Matrix([X, Y]) + Q): _a = (X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t _b = (X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t Xt = S(_a + _b)/2 Yt = S(_a - _b)/(2*sqrt(D)) s = P*Matrix([Xt, Yt]) + Q sol.add(tuple(s)) X, Y = X*T + D*U*Y, X*U + Y*T return sol def is_solution_quad(var, coeff, u, v): """ Check whether `(u, v)` is solution to the quadratic binary diophantine equation with the variable list ``var`` and coefficient dictionary ``coeff``. Not intended for use by normal users. """ reps = dict(zip(var, (u, v))) eq = Add(*[j*i.xreplace(reps) for i, j in coeff.items()]) return _mexpand(eq) == 0
[docs]def diop_DN(D, N, t=symbols("t", integer=True)): """ Solves the equation `x^2 - Dy^2 = N`. Mainly concerned with the case `D > 0, D` is not a perfect square, which is the same as the generalized Pell equation. The LMM algorithm [1]_ is used to solve this equation. Returns one solution tuple, (`x, y)` for each class of the solutions. Other solutions of the class can be constructed according to the values of ``D`` and ``N``. Usage ===== ``diop_DN(D, N, t)``: D and N are integers as in `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. Details ======= ``D`` and ``N`` correspond to D and N in the equation. ``t`` is the parameter to be used in the solutions. Examples ======== >>> from sympy.solvers.diophantine import diop_DN >>> diop_DN(13, -4) # Solves equation x**2 - 13*y**2 = -4 [(3, 1), (393, 109), (36, 10)] The output can be interpreted as follows: There are three fundamental solutions to the equation `x^2 - 13y^2 = -4` given by (3, 1), (393, 109) and (36, 10). Each tuple is in the form (x, y), i.e. solution (3, 1) means that `x = 3` and `y = 1`. >>> diop_DN(986, 1) # Solves equation x**2 - 986*y**2 = 1 [(49299, 1570)] See Also ======== find_DN(), diop_bf_DN() References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Pages 16 - 17. [online], Available: http://www.jpr2718.org/pell.pdf """ if D < 0: if N == 0: return [(0, 0)] elif N < 0: return [] elif N > 0: sol = [] for d in divisors(square_factor(N)): sols = cornacchia(1, -D, N // d**2) if sols: for x, y in sols: sol.append((d*x, d*y)) if D == -1: sol.append((d*y, d*x)) return sol elif D == 0: if N < 0: return [] if N == 0: return [(0, t)] sN, _exact = integer_nthroot(N, 2) if _exact: return [(sN, t)] else: return [] else: # D > 0 sD, _exact = integer_nthroot(D, 2) if _exact: if N == 0: return [(sD*t, t)] else: sol = [] for y in range(floor(sign(N)*(N - 1)/(2*sD)) + 1): try: sq, _exact = integer_nthroot(D*y**2 + N, 2) except ValueError: _exact = False if _exact: sol.append((sq, y)) return sol elif 1 < N**2 < D: # It is much faster to call `_special_diop_DN`. return _special_diop_DN(D, N) else: if N == 0: return [(0, 0)] elif abs(N) == 1: pqa = PQa(0, 1, D) j = 0 G = [] B = [] for i in pqa: a = i[2] G.append(i[5]) B.append(i[4]) if j != 0 and a == 2*sD: break j = j + 1 if _odd(j): if N == -1: x = G[j - 1] y = B[j - 1] else: count = j while count < 2*j - 1: i = next(pqa) G.append(i[5]) B.append(i[4]) count += 1 x = G[count] y = B[count] else: if N == 1: x = G[j - 1] y = B[j - 1] else: return [] return [(x, y)] else: fs = [] sol = [] div = divisors(N) for d in div: if divisible(N, d**2): fs.append(d) for f in fs: m = N // f**2 zs = sqrt_mod(D, abs(m), all_roots=True) zs = [i for i in zs if i <= abs(m) // 2 ] if abs(m) != 2: zs = zs + [-i for i in zs if i] # omit dupl 0 for z in zs: pqa = PQa(z, abs(m), D) j = 0 G = [] B = [] for i in pqa: a = i[2] G.append(i[5]) B.append(i[4]) if j != 0 and abs(i[1]) == 1: r = G[j-1] s = B[j-1] if r**2 - D*s**2 == m: sol.append((f*r, f*s)) elif diop_DN(D, -1) != []: a = diop_DN(D, -1) sol.append((f*(r*a[0][0] + a[0][1]*s*D), f*(r*a[0][1] + s*a[0][0]))) break j = j + 1 if j == length(z, abs(m), D): break return sol
def _special_diop_DN(D, N): """ Solves the equation `x^2 - Dy^2 = N` for the special case where `1 < N**2 < D` and `D` is not a perfect square. It is better to call `diop_DN` rather than this function, as the former checks the condition `1 < N**2 < D`, and calls the latter only if appropriate. Usage ===== WARNING: Internal method. Do not call directly! ``_special_diop_DN(D, N)``: D and N are integers as in `x^2 - Dy^2 = N`. Details ======= ``D`` and ``N`` correspond to D and N in the equation. Examples ======== >>> from sympy.solvers.diophantine import _special_diop_DN >>> _special_diop_DN(13, -3) # Solves equation x**2 - 13*y**2 = -3 [(7, 2), (137, 38)] The output can be interpreted as follows: There are two fundamental solutions to the equation `x^2 - 13y^2 = -3` given by (7, 2) and (137, 38). Each tuple is in the form (x, y), i.e. solution (7, 2) means that `x = 7` and `y = 2`. >>> _special_diop_DN(2445, -20) # Solves equation x**2 - 2445*y**2 = -20 [(445, 9), (17625560, 356454), (698095554475, 14118073569)] See Also ======== diop_DN() References ========== .. [1] Section 4.4.4 of the following book: Quadratic Diophantine Equations, T. Andreescu and D. Andrica, Springer, 2015. """ # The following assertion was removed for efficiency, with the understanding # that this method is not called directly. The parent method, `diop_DN` # is responsible for performing the appropriate checks. # # assert (1 < N**2 < D) and (not integer_nthroot(D, 2)[1]) sqrt_D = sqrt(D) F = [(N, 1)] f = 2 while True: f2 = f**2 if f2 > abs(N): break n, r = divmod(N, f2) if r == 0: F.append((n, f)) f += 1 P = 0 Q = 1 G0, G1 = 0, 1 B0, B1 = 1, 0 solutions = [] i = 0 while True: a = floor((P + sqrt_D) / Q) P = a*Q - P Q = (D - P**2) // Q G2 = a*G1 + G0 B2 = a*B1 + B0 for n, f in F: if G2**2 - D*B2**2 == n: solutions.append((f*G2, f*B2)) i += 1 if Q == 1 and i % 2 == 0: break G0, G1 = G1, G2 B0, B1 = B1, B2 return solutions
[docs]def cornacchia(a, b, m): r""" Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`. Uses the algorithm due to Cornacchia. The method only finds primitive solutions, i.e. ones with `\gcd(x, y) = 1`. So this method can't be used to find the solutions of `x^2 + y^2 = 20` since the only solution to former is `(x, y) = (4, 2)` and it is not primitive. When `a = b`, only the solutions with `x \leq y` are found. For more details, see the References. Examples ======== >>> from sympy.solvers.diophantine import cornacchia >>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35 {(2, 3), (4, 1)} >>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25 {(4, 3)} References =========== .. [1] A. Nitaj, "L'algorithme de Cornacchia" .. [2] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia's method, [online], Available: http://www.numbertheory.org/php/cornacchia.html See Also ======== sympy.utilities.iterables.signed_permutations """ sols = set() a1 = igcdex(a, m)[0] v = sqrt_mod(-b*a1, m, all_roots=True) if not v: return None for t in v: if t < m // 2: continue u, r = t, m while True: u, r = r, u % r if a*r**2 < m: break m1 = m - a*r**2 if m1 % b == 0: m1 = m1 // b s, _exact = integer_nthroot(m1, 2) if _exact: if a == b and r < s: r, s = s, r sols.add((int(r), int(s))) return sols
[docs]def PQa(P_0, Q_0, D): r""" Returns useful information needed to solve the Pell equation. There are six sequences of integers defined related to the continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`, namely {`P_{i}`}, {`Q_{i}`}, {`a_{i}`},{`A_{i}`}, {`B_{i}`}, {`G_{i}`}. ``PQa()`` Returns these values as a 6-tuple in the same order as mentioned above. Refer [1]_ for more detailed information. Usage ===== ``PQa(P_0, Q_0, D)``: ``P_0``, ``Q_0`` and ``D`` are integers corresponding to `P_{0}`, `Q_{0}` and `D` in the continued fraction `\\frac{P_{0} + \sqrt{D}}{Q_{0}}`. Also it's assumed that `P_{0}^2 == D mod(|Q_{0}|)` and `D` is square free. Examples ======== >>> from sympy.solvers.diophantine import PQa >>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4 >>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0) (13, 4, 3, 3, 1, -1) >>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1) (-1, 1, 1, 4, 1, 3) References ========== .. [1] Solving the generalized Pell equation x^2 - Dy^2 = N, John P. Robertson, July 31, 2004, Pages 4 - 8. http://www.jpr2718.org/pell.pdf """ A_i_2 = B_i_1 = 0 A_i_1 = B_i_2 = 1 G_i_2 = -P_0 G_i_1 = Q_0 P_i = P_0 Q_i = Q_0 while(1): a_i = floor((P_i + sqrt(D))/Q_i) A_i = a_i*A_i_1 + A_i_2 B_i = a_i*B_i_1 + B_i_2 G_i = a_i*G_i_1 + G_i_2 yield P_i, Q_i, a_i, A_i, B_i, G_i A_i_1, A_i_2 = A_i, A_i_1 B_i_1, B_i_2 = B_i, B_i_1 G_i_1, G_i_2 = G_i, G_i_1 P_i = a_i*Q_i - P_i Q_i = (D - P_i**2)/Q_i
[docs]def diop_bf_DN(D, N, t=symbols("t", integer=True)): r""" Uses brute force to solve the equation, `x^2 - Dy^2 = N`. Mainly concerned with the generalized Pell equation which is the case when `D > 0, D` is not a perfect square. For more information on the case refer [1]_. Let `(t, u)` be the minimal positive solution of the equation `x^2 - Dy^2 = 1`. Then this method requires `\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}` to be small. Usage ===== ``diop_bf_DN(D, N, t)``: ``D`` and ``N`` are coefficients in `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. Details ======= ``D`` and ``N`` correspond to D and N in the equation. ``t`` is the parameter to be used in the solutions. Examples ======== >>> from sympy.solvers.diophantine import diop_bf_DN >>> diop_bf_DN(13, -4) [(3, 1), (-3, 1), (36, 10)] >>> diop_bf_DN(986, 1) [(49299, 1570)] See Also ======== diop_DN() References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Page 15. http://www.jpr2718.org/pell.pdf """ D = as_int(D) N = as_int(N) sol = [] a = diop_DN(D, 1) u = a[0][0] v = a[0][1] if abs(N) == 1: return diop_DN(D, N) elif N > 1: L1 = 0 L2 = integer_nthroot(int(N*(u - 1)/(2*D)), 2)[0] + 1 elif N < -1: L1, _exact = integer_nthroot(-int(N/D), 2) if not _exact: L1 += 1 L2 = integer_nthroot(-int(N*(u + 1)/(2*D)), 2)[0] + 1 else: # N = 0 if D < 0: return [(0, 0)] elif D == 0: return [(0, t)] else: sD, _exact = integer_nthroot(D, 2) if _exact: return [(sD*t, t), (-sD*t, t)] else: return [(0, 0)] for y in range(L1, L2): try: x, _exact = integer_nthroot(N + D*y**2, 2) except ValueError: _exact = False if _exact: sol.append((x, y)) if not equivalent(x, y, -x, y, D, N): sol.append((-x, y)) return sol
[docs]def equivalent(u, v, r, s, D, N): """ Returns True if two solutions `(u, v)` and `(r, s)` of `x^2 - Dy^2 = N` belongs to the same equivalence class and False otherwise. Two solutions `(u, v)` and `(r, s)` to the above equation fall to the same equivalence class iff both `(ur - Dvs)` and `(us - vr)` are divisible by `N`. See reference [1]_. No test is performed to test whether `(u, v)` and `(r, s)` are actually solutions to the equation. User should take care of this. Usage ===== ``equivalent(u, v, r, s, D, N)``: `(u, v)` and `(r, s)` are two solutions of the equation `x^2 - Dy^2 = N` and all parameters involved are integers. Examples ======== >>> from sympy.solvers.diophantine import equivalent >>> equivalent(18, 5, -18, -5, 13, -1) True >>> equivalent(3, 1, -18, 393, 109, -4) False References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Page 12. http://www.jpr2718.org/pell.pdf """ return divisible(u*r - D*v*s, N) and divisible(u*s - v*r, N)
def length(P, Q, D): r""" Returns the (length of aperiodic part + length of periodic part) of continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`. It is important to remember that this does NOT return the length of the periodic part but the sum of the lengths of the two parts as mentioned above. Usage ===== ``length(P, Q, D)``: ``P``, ``Q`` and ``D`` are integers corresponding to the continued fraction `\\frac{P + \sqrt{D}}{Q}`. Details ======= ``P``, ``D`` and ``Q`` corresponds to P, D and Q in the continued fraction, `\\frac{P + \sqrt{D}}{Q}`. Examples ======== >>> from sympy.solvers.diophantine import length >>> length(-2 , 4, 5) # (-2 + sqrt(5))/4 3 >>> length(-5, 4, 17) # (-5 + sqrt(17))/4 5 See Also ======== sympy.ntheory.continued_fraction.continued_fraction_periodic """ from sympy.ntheory.continued_fraction import continued_fraction_periodic v = continued_fraction_periodic(P, Q, D) if type(v[-1]) is list: rpt = len(v[-1]) nonrpt = len(v) - 1 else: rpt = 0 nonrpt = len(v) return rpt + nonrpt
[docs]def transformation_to_DN(eq): """ This function transforms general quadratic, `ax^2 + bxy + cy^2 + dx + ey + f = 0` to more easy to deal with `X^2 - DY^2 = N` form. This is used to solve the general quadratic equation by transforming it to the latter form. Refer [1]_ for more detailed information on the transformation. This function returns a tuple (A, B) where A is a 2 X 2 matrix and B is a 2 X 1 matrix such that, Transpose([x y]) = A * Transpose([X Y]) + B Usage ===== ``transformation_to_DN(eq)``: where ``eq`` is the quadratic to be transformed. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine import transformation_to_DN >>> from sympy.solvers.diophantine import classify_diop >>> A, B = transformation_to_DN(x**2 - 3*x*y - y**2 - 2*y + 1) >>> A Matrix([ [1/26, 3/26], [ 0, 1/13]]) >>> B Matrix([ [-6/13], [-4/13]]) A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B. Substituting these values for `x` and `y` and a bit of simplifying work will give an equation of the form `x^2 - Dy^2 = N`. >>> from sympy.abc import X, Y >>> from sympy import Matrix, simplify >>> u = (A*Matrix([X, Y]) + B)[0] # Transformation for x >>> u X/26 + 3*Y/26 - 6/13 >>> v = (A*Matrix([X, Y]) + B)[1] # Transformation for y >>> v Y/13 - 4/13 Next we will substitute these formulas for `x` and `y` and do ``simplify()``. >>> eq = simplify((x**2 - 3*x*y - y**2 - 2*y + 1).subs(zip((x, y), (u, v)))) >>> eq X**2/676 - Y**2/52 + 17/13 By multiplying the denominator appropriately, we can get a Pell equation in the standard form. >>> eq * 676 X**2 - 13*Y**2 + 884 If only the final equation is needed, ``find_DN()`` can be used. See Also ======== find_DN() References ========== .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, John P.Robertson, May 8, 2003, Page 7 - 11. http://www.jpr2718.org/ax2p.pdf """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "binary_quadratic": return _transformation_to_DN(var, coeff)
def _transformation_to_DN(var, coeff): x, y = var a = coeff[x**2] b = coeff[x*y] c = coeff[y**2] d = coeff[x] e = coeff[y] f = coeff[1] a, b, c, d, e, f = [as_int(i) for i in _remove_gcd(a, b, c, d, e, f)] X, Y = symbols("X, Y", integer=True) if b: B, C = _rational_pq(2*a, b) A, T = _rational_pq(a, B**2) # eq_1 = A*B*X**2 + B*(c*T - A*C**2)*Y**2 + d*T*X + (B*e*T - d*T*C)*Y + f*T*B coeff = {X**2: A*B, X*Y: 0, Y**2: B*(c*T - A*C**2), X: d*T, Y: B*e*T - d*T*C, 1: f*T*B} A_0, B_0 = _transformation_to_DN([X, Y], coeff) return Matrix(2, 2, [S(1)/B, -S(C)/B, 0, 1])*A_0, Matrix(2, 2, [S(1)/B, -S(C)/B, 0, 1])*B_0 else: if d: B, C = _rational_pq(2*a, d) A, T = _rational_pq(a, B**2) # eq_2 = A*X**2 + c*T*Y**2 + e*T*Y + f*T - A*C**2 coeff = {X**2: A, X*Y: 0, Y**2: c*T, X: 0, Y: e*T, 1: f*T - A*C**2} A_0, B_0 = _transformation_to_DN([X, Y], coeff) return Matrix(2, 2, [S(1)/B, 0, 0, 1])*A_0, Matrix(2, 2, [S(1)/B, 0, 0, 1])*B_0 + Matrix([-S(C)/B, 0]) else: if e: B, C = _rational_pq(2*c, e) A, T = _rational_pq(c, B**2) # eq_3 = a*T*X**2 + A*Y**2 + f*T - A*C**2 coeff = {X**2: a*T, X*Y: 0, Y**2: A, X: 0, Y: 0, 1: f*T - A*C**2} A_0, B_0 = _transformation_to_DN([X, Y], coeff) return Matrix(2, 2, [1, 0, 0, S(1)/B])*A_0, Matrix(2, 2, [1, 0, 0, S(1)/B])*B_0 + Matrix([0, -S(C)/B]) else: # TODO: pre-simplification: Not necessary but may simplify # the equation. return Matrix(2, 2, [S(1)/a, 0, 0, 1]), Matrix([0, 0])
[docs]def find_DN(eq): """ This function returns a tuple, `(D, N)` of the simplified form, `x^2 - Dy^2 = N`, corresponding to the general quadratic, `ax^2 + bxy + cy^2 + dx + ey + f = 0`. Solving the general quadratic is then equivalent to solving the equation `X^2 - DY^2 = N` and transforming the solutions by using the transformation matrices returned by ``transformation_to_DN()``. Usage ===== ``find_DN(eq)``: where ``eq`` is the quadratic to be transformed. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine import find_DN >>> find_DN(x**2 - 3*x*y - y**2 - 2*y + 1) (13, -884) Interpretation of the output is that we get `X^2 -13Y^2 = -884` after transforming `x^2 - 3xy - y^2 - 2y + 1` using the transformation returned by ``transformation_to_DN()``. See Also ======== transformation_to_DN() References ========== .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, John P.Robertson, May 8, 2003, Page 7 - 11. http://www.jpr2718.org/ax2p.pdf """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "binary_quadratic": return _find_DN(var, coeff)
def _find_DN(var, coeff): x, y = var X, Y = symbols("X, Y", integer=True) A, B = _transformation_to_DN(var, coeff) u = (A*Matrix([X, Y]) + B)[0] v = (A*Matrix([X, Y]) + B)[1] eq = x**2*coeff[x**2] + x*y*coeff[x*y] + y**2*coeff[y**2] + x*coeff[x] + y*coeff[y] + coeff[1] simplified = _mexpand(eq.subs(zip((x, y), (u, v)))) coeff = simplified.as_coefficients_dict() return -coeff[Y**2]/coeff[X**2], -coeff[1]/coeff[X**2] def check_param(x, y, a, t): """ If there is a number modulo ``a`` such that ``x`` and ``y`` are both integers, then return a parametric representation for ``x`` and ``y`` else return (None, None). Here ``x`` and ``y`` are functions of ``t``. """ from sympy.simplify.simplify import clear_coefficients if x.is_number and not x.is_Integer: return (None, None) if y.is_number and not y.is_Integer: return (None, None) m, n = symbols("m, n", integer=True) c, p = (m*x + n*y).as_content_primitive() if a % c.q: return (None, None) # clear_coefficients(mx + b, R)[1] -> (R - b)/m eq = clear_coefficients(x, m)[1] - clear_coefficients(y, n)[1] junk, eq = eq.as_content_primitive() return diop_solve(eq, t)
[docs]def diop_ternary_quadratic(eq): """ Solves the general quadratic ternary form, `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. Returns a tuple `(x, y, z)` which is a base solution for the above equation. If there are no solutions, `(None, None, None)` is returned. Usage ===== ``diop_ternary_quadratic(eq)``: Return a tuple containing a basic solution to ``eq``. Details ======= ``eq`` should be an homogeneous expression of degree two in three variables and it is assumed to be zero. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine import diop_ternary_quadratic >>> diop_ternary_quadratic(x**2 + 3*y**2 - z**2) (1, 0, 1) >>> diop_ternary_quadratic(4*x**2 + 5*y**2 - z**2) (1, 0, 2) >>> diop_ternary_quadratic(45*x**2 - 7*y**2 - 8*x*y - z**2) (28, 45, 105) >>> diop_ternary_quadratic(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y) (9, 1, 5) """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type in ( "homogeneous_ternary_quadratic", "homogeneous_ternary_quadratic_normal"): return _diop_ternary_quadratic(var, coeff)
def _diop_ternary_quadratic(_var, coeff): x, y, z = _var var = [x, y, z] # Equations of the form B*x*y + C*z*x + E*y*z = 0 and At least two of the # coefficients A, B, C are non-zero. # There are infinitely many solutions for the equation. # Ex: (0, 0, t), (0, t, 0), (t, 0, 0) # Equation can be re-written as y*(B*x + E*z) = -C*x*z and we can find rather # unobvious solutions. Set y = -C and B*x + E*z = x*z. The latter can be solved by # using methods for binary quadratic diophantine equations. Let's select the # solution which minimizes |x| + |z| if not any(coeff[i**2] for i in var): if coeff[x*z]: sols = diophantine(coeff[x*y]*x + coeff[y*z]*z - x*z) s = sols.pop() min_sum = abs(s[0]) + abs(s[1]) for r in sols: if abs(r[0]) + abs(r[1]) < min_sum: s = r min_sum = abs(s[0]) + abs(s[1]) x_0, y_0, z_0 = s[0], -coeff[x*z], s[1] else: var[0], var[1] = _var[1], _var[0] y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff) return _remove_gcd(x_0, y_0, z_0) if coeff[x**2] == 0: # If the coefficient of x is zero change the variables if coeff[y**2] == 0: var[0], var[2] = _var[2], _var[0] z_0, y_0, x_0 = _diop_ternary_quadratic(var, coeff) else: var[0], var[1] = _var[1], _var[0] y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff) else: if coeff[x*y] or coeff[x*z]: # Apply the transformation x --> X - (B*y + C*z)/(2*A) A = coeff[x**2] B = coeff[x*y] C = coeff[x*z] D = coeff[y**2] E = coeff[y*z] F = coeff[z**2] _coeff = dict() _coeff[x**2] = 4*A**2 _coeff[y**2] = 4*A*D - B**2 _coeff[z**2] = 4*A*F - C**2 _coeff[y*z] = 4*A*E - 2*B*C _coeff[x*y] = 0 _coeff[x*z] = 0 x_0, y_0, z_0 = _diop_ternary_quadratic(var, _coeff) if x_0 is None: return (None, None, None) p, q = _rational_pq(B*y_0 + C*z_0, 2*A) x_0, y_0, z_0 = x_0*q - p, y_0*q, z_0*q elif coeff[z*y] != 0: if coeff[y**2] == 0: if coeff[z**2] == 0: # Equations of the form A*x**2 + E*yz = 0. A = coeff[x**2] E = coeff[y*z] b, a = _rational_pq(-E, A) x_0, y_0, z_0 = b, a, b else: # Ax**2 + E*y*z + F*z**2 = 0 var[0], var[2] = _var[2], _var[0] z_0, y_0, x_0 = _diop_ternary_quadratic(var, coeff) else: # A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, C may be zero var[0], var[1] = _var[1], _var[0] y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff) else: # Ax**2 + D*y**2 + F*z**2 = 0, C may be zero x_0, y_0, z_0 = _diop_ternary_quadratic_normal(var, coeff) return _remove_gcd(x_0, y_0, z_0) def transformation_to_normal(eq): """ Returns the transformation Matrix that converts a general ternary quadratic equation `eq` (`ax^2 + by^2 + cz^2 + dxy + eyz + fxz`) to a form without cross terms: `ax^2 + by^2 + cz^2 = 0`. This is not used in solving ternary quadratics; it is only implemented for the sake of completeness. """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type in ( "homogeneous_ternary_quadratic", "homogeneous_ternary_quadratic_normal"): return _transformation_to_normal(var, coeff) def _transformation_to_normal(var, coeff): _var = list(var) # copy x, y, z = var if not any(coeff[i**2] for i in var): # https://math.stackexchange.com/questions/448051/transform-quadratic-ternary-form-to-normal-form/448065#448065 a = coeff[x*y] b = coeff[y*z] c = coeff[x*z] swap = False if not a: # b can't be 0 or else there aren't 3 vars swap = True a, b = b, a T = Matrix(((1, 1, -b/a), (1, -1, -c/a), (0, 0, 1))) if swap: T.row_swap(0, 1) T.col_swap(0, 1) return T if coeff[x**2] == 0: # If the coefficient of x is zero change the variables if coeff[y**2] == 0: _var[0], _var[2] = var[2], var[0] T = _transformation_to_normal(_var, coeff) T.row_swap(0, 2) T.col_swap(0, 2) return T else: _var[0], _var[1] = var[1], var[0] T = _transformation_to_normal(_var, coeff) T.row_swap(0, 1) T.col_swap(0, 1) return T # Apply the transformation x --> X - (B*Y + C*Z)/(2*A) if coeff[x*y] != 0 or coeff[x*z] != 0: A = coeff[x**2] B = coeff[x*y] C = coeff[x*z] D = coeff[y**2] E = coeff[y*z] F = coeff[z**2] _coeff = dict() _coeff[x**2] = 4*A**2 _coeff[y**2] = 4*A*D - B**2 _coeff[z**2] = 4*A*F - C**2 _coeff[y*z] = 4*A*E - 2*B*C _coeff[x*y] = 0 _coeff[x*z] = 0 T_0 = _transformation_to_normal(_var, _coeff) return Matrix(3, 3, [1, S(-B)/(2*A), S(-C)/(2*A), 0, 1, 0, 0, 0, 1])*T_0 elif coeff[y*z] != 0: if coeff[y**2] == 0: if coeff[z**2] == 0: # Equations of the form A*x**2 + E*yz = 0. # Apply transformation y -> Y + Z ans z -> Y - Z return Matrix(3, 3, [1, 0, 0, 0, 1, 1, 0, 1, -1]) else: # Ax**2 + E*y*z + F*z**2 = 0 _var[0], _var[2] = var[2], var[0] T = _transformation_to_normal(_var, coeff) T.row_swap(0, 2) T.col_swap(0, 2) return T else: # A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, F may be zero _var[0], _var[1] = var[1], var[0] T = _transformation_to_normal(_var, coeff) T.row_swap(0, 1) T.col_swap(0, 1) return T else: return Matrix.eye(3)
[docs]def parametrize_ternary_quadratic(eq): """ Returns the parametrized general solution for the ternary quadratic equation ``eq`` which has the form `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine import parametrize_ternary_quadratic >>> parametrize_ternary_quadratic(x**2 + y**2 - z**2) (2*p*q, p**2 - q**2, p**2 + q**2) Here `p` and `q` are two co-prime integers. >>> parametrize_ternary_quadratic(3*x**2 + 2*y**2 - z**2 - 2*x*y + 5*y*z - 7*y*z) (2*p**2 - 2*p*q - q**2, 2*p**2 + 2*p*q - q**2, 2*p**2 - 2*p*q + 3*q**2) >>> parametrize_ternary_quadratic(124*x**2 - 30*y**2 - 7729*z**2) (-1410*p**2 - 363263*q**2, 2700*p**2 + 30916*p*q - 695610*q**2, -60*p**2 + 5400*p*q + 15458*q**2) References ========== .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998. """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type in ( "homogeneous_ternary_quadratic", "homogeneous_ternary_quadratic_normal"): x_0, y_0, z_0 = _diop_ternary_quadratic(var, coeff) return _parametrize_ternary_quadratic( (x_0, y_0, z_0), var, coeff)
def _parametrize_ternary_quadratic(solution, _var, coeff): # called for a*x**2 + b*y**2 + c*z**2 + d*x*y + e*y*z + f*x*z = 0 assert 1 not in coeff x, y, z = _var x_0, y_0, z_0 = solution v = list(_var) # copy if x_0 is None: return (None, None, None) if solution.count(0) >= 2: # if there are 2 zeros the equation reduces # to k*X**2 == 0 where X is x, y, or z so X must # be zero, too. So there is only the trivial # solution. return (None, None, None) if x_0 == 0: v[0], v[1] = v[1], v[0] y_p, x_p, z_p = _parametrize_ternary_quadratic( (y_0, x_0, z_0), v, coeff) return x_p, y_p, z_p x, y, z = v r, p, q = symbols("r, p, q", integer=True) eq = sum(k*v for k, v in coeff.items()) eq_1 = _mexpand(eq.subs(zip( (x, y, z), (r*x_0, r*y_0 + p, r*z_0 + q)))) A, B = eq_1.as_independent(r, as_Add=True) x = A*x_0 y = (A*y_0 - _mexpand(B/r*p)) z = (A*z_0 - _mexpand(B/r*q)) return x, y, z
[docs]def diop_ternary_quadratic_normal(eq): """ Solves the quadratic ternary diophantine equation, `ax^2 + by^2 + cz^2 = 0`. Here the coefficients `a`, `b`, and `c` should be non zero. Otherwise the equation will be a quadratic binary or univariate equation. If solvable, returns a tuple `(x, y, z)` that satisfies the given equation. If the equation does not have integer solutions, `(None, None, None)` is returned. Usage ===== ``diop_ternary_quadratic_normal(eq)``: where ``eq`` is an equation of the form `ax^2 + by^2 + cz^2 = 0`. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine import diop_ternary_quadratic_normal >>> diop_ternary_quadratic_normal(x**2 + 3*y**2 - z**2) (1, 0, 1) >>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2) (1, 0, 2) >>> diop_ternary_quadratic_normal(34*x**2 - 3*y**2 - 301*z**2) (4, 9, 1) """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "homogeneous_ternary_quadratic_normal": return _diop_ternary_quadratic_normal(var, coeff)
def _diop_ternary_quadratic_normal(var, coeff): x, y, z = var a = coeff[x**2] b = coeff[y**2] c = coeff[z**2] try: assert len([k for k in coeff if coeff[k]]) == 3 assert all(coeff[i**2] for i in var) except AssertionError: raise ValueError(filldedent(''' coeff dict is not consistent with assumption of this routine: coefficients should be those of an expression in the form a*x**2 + b*y**2 + c*z**2 where a*b*c != 0.''')) (sqf_of_a, sqf_of_b, sqf_of_c), (a_1, b_1, c_1), (a_2, b_2, c_2) = \ sqf_normal(a, b, c, steps=True) A = -a_2*c_2 B = -b_2*c_2 # If following two conditions are satisified then there are no solutions if A < 0 and B < 0: return (None, None, None) if ( sqrt_mod(-b_2*c_2, a_2) is None or sqrt_mod(-c_2*a_2, b_2) is None or sqrt_mod(-a_2*b_2, c_2) is None): return (None, None, None) z_0, x_0, y_0 = descent(A, B) z_0, q = _rational_pq(z_0, abs(c_2)) x_0 *= q y_0 *= q x_0, y_0, z_0 = _remove_gcd(x_0, y_0, z_0) # Holzer reduction if sign(a) == sign(b): x_0, y_0, z_0 = holzer(x_0, y_0, z_0, abs(a_2), abs(b_2), abs(c_2)) elif sign(a) == sign(c): x_0, z_0, y_0 = holzer(x_0, z_0, y_0, abs(a_2), abs(c_2), abs(b_2)) else: y_0, z_0, x_0 = holzer(y_0, z_0, x_0, abs(b_2), abs(c_2), abs(a_2)) x_0 = reconstruct(b_1, c_1, x_0) y_0 = reconstruct(a_1, c_1, y_0) z_0 = reconstruct(a_1, b_1, z_0) sq_lcm = ilcm(sqf_of_a, sqf_of_b, sqf_of_c) x_0 = abs(x_0*sq_lcm//sqf_of_a) y_0 = abs(y_0*sq_lcm//sqf_of_b) z_0 = abs(z_0*sq_lcm//sqf_of_c) return _remove_gcd(x_0, y_0, z_0)
[docs]def sqf_normal(a, b, c, steps=False): """ Return `a', b', c'`, the coefficients of the square-free normal form of `ax^2 + by^2 + cz^2 = 0`, where `a', b', c'` are pairwise prime. If `steps` is True then also return three tuples: `sq`, `sqf`, and `(a', b', c')` where `sq` contains the square factors of `a`, `b` and `c` after removing the `gcd(a, b, c)`; `sqf` contains the values of `a`, `b` and `c` after removing both the `gcd(a, b, c)` and the square factors. The solutions for `ax^2 + by^2 + cz^2 = 0` can be recovered from the solutions of `a'x^2 + b'y^2 + c'z^2 = 0`. Examples ======== >>> from sympy.solvers.diophantine import sqf_normal >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11) (11, 1, 5) >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11, True) ((3, 1, 7), (5, 55, 11), (11, 1, 5)) References ========== .. [1] Legendre's Theorem, Legrange's Descent, http://public.csusm.edu/aitken_html/notes/legendre.pdf See Also ======== reconstruct() """ ABC = A, B, C = _remove_gcd(a, b, c) sq = tuple(square_factor(i) for i in ABC) sqf = A, B, C = tuple([i//j**2 for i,j in zip(ABC, sq)]) pc = igcd(A, B) A /= pc B /= pc pa = igcd(B, C) B /= pa C /= pa pb = igcd(A, C) A /= pb B /= pb A *= pa B *= pb C *= pc if steps: return (sq, sqf, (A, B, C)) else: return A, B, C
[docs]def square_factor(a): r""" Returns an integer `c` s.t. `a = c^2k, \ c,k \in Z`. Here `k` is square free. `a` can be given as an integer or a dictionary of factors. Examples ======== >>> from sympy.solvers.diophantine import square_factor >>> square_factor(24) 2 >>> square_factor(-36*3) 6 >>> square_factor(1) 1 >>> square_factor({3: 2, 2: 1, -1: 1}) # -18 3 See Also ======== sympy.ntheory.factor_.core """ f = a if isinstance(a, dict) else factorint(a) return Mul(*[p**(e//2) for p, e in f.items()])
[docs]def reconstruct(A, B, z): """ Reconstruct the `z` value of an equivalent solution of `ax^2 + by^2 + cz^2` from the `z` value of a solution of the square-free normal form of the equation, `a'*x^2 + b'*y^2 + c'*z^2`, where `a'`, `b'` and `c'` are square free and `gcd(a', b', c') == 1`. """ f = factorint(igcd(A, B)) for p, e in f.items(): if e != 1: raise ValueError('a and b should be square-free') z *= p return z
[docs]def ldescent(A, B): """ Return a non-trivial solution to `w^2 = Ax^2 + By^2` using Lagrange's method; return None if there is no such solution. . Here, `A \\neq 0` and `B \\neq 0` and `A` and `B` are square free. Output a tuple `(w_0, x_0, y_0)` which is a solution to the above equation. Examples ======== >>> from sympy.solvers.diophantine import ldescent >>> ldescent(1, 1) # w^2 = x^2 + y^2 (1, 1, 0) >>> ldescent(4, -7) # w^2 = 4x^2 - 7y^2 (2, -1, 0) This means that `x = -1, y = 0` and `w = 2` is a solution to the equation `w^2 = 4x^2 - 7y^2` >>> ldescent(5, -1) # w^2 = 5x^2 - y^2 (2, 1, -1) References ========== .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998. .. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, [online], Available: http://eprints.nottingham.ac.uk/60/1/kvxefz87.pdf """ if abs(A) > abs(B): w, y, x = ldescent(B, A) return w, x, y if A == 1: return (1, 1, 0) if B == 1: return (1, 0, 1) if B == -1: # and A == -1 return r = sqrt_mod(A, B) Q = (r**2 - A) // B if Q == 0: B_0 = 1 d = 0 else: div = divisors(Q) B_0 = None for i in div: sQ, _exact = integer_nthroot(abs(Q) // i, 2) if _exact: B_0, d = sign(Q)*i, sQ break if B_0 is not None: W, X, Y = ldescent(A, B_0) return _remove_gcd((-A*X + r*W), (r*X - W), Y*(B_0*d))
[docs]def descent(A, B): """ Returns a non-trivial solution, (x, y, z), to `x^2 = Ay^2 + Bz^2` using Lagrange's descent method with lattice-reduction. `A` and `B` are assumed to be valid for such a solution to exist. This is faster than the normal Lagrange's descent algorithm because the Gaussian reduction is used. Examples ======== >>> from sympy.solvers.diophantine import descent >>> descent(3, 1) # x**2 = 3*y**2 + z**2 (1, 0, 1) `(x, y, z) = (1, 0, 1)` is a solution to the above equation. >>> descent(41, -113) (-16, -3, 1) References ========== .. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0. """ if abs(A) > abs(B): x, y, z = descent(B, A) return x, z, y if B == 1: return (1, 0, 1) if A == 1: return (1, 1, 0) if B == -A: return (0, 1, 1) if B == A: x, z, y = descent(-1, A) return (A*y, z, x) w = sqrt_mod(A, B) x_0, z_0 = gaussian_reduce(w, A, B) t = (x_0**2 - A*z_0**2) // B t_2 = square_factor(t) t_1 = t // t_2**2 x_1, z_1, y_1 = descent(A, t_1) return _remove_gcd(x_0*x_1 + A*z_0*z_1, z_0*x_1 + x_0*z_1, t_1*t_2*y_1)
[docs]def gaussian_reduce(w, a, b): r""" Returns a reduced solution `(x, z)` to the congruence `X^2 - aZ^2 \equiv 0 \ (mod \ b)` so that `x^2 + |a|z^2` is minimal. Details ======= Here ``w`` is a solution of the congruence `x^2 \equiv a \ (mod \ b)` References ========== .. [1] Gaussian lattice Reduction [online]. Available: http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404 .. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0. """ u = (0, 1) v = (1, 0) if dot(u, v, w, a, b) < 0: v = (-v[0], -v[1]) if norm(u, w, a, b) < norm(v, w, a, b): u, v = v, u while norm(u, w, a, b) > norm(v, w, a, b): k = dot(u, v, w, a, b) // dot(v, v, w, a, b) u, v = v, (u[0]- k*v[0], u[1]- k*v[1]) u, v = v, u if dot(u, v, w, a, b) < dot(v, v, w, a, b)/2 or norm((u[0]-v[0], u[1]-v[1]), w, a, b) > norm(v, w, a, b): c = v else: c = (u[0] - v[0], u[1] - v[1]) return c[0]*w + b*c[1], c[0]
def dot(u, v, w, a, b): r""" Returns a special dot product of the vectors `u = (u_{1}, u_{2})` and `v = (v_{1}, v_{2})` which is defined in order to reduce solution of the congruence equation `X^2 - aZ^2 \equiv 0 \ (mod \ b)`. """ u_1, u_2 = u v_1, v_2 = v return (w*u_1 + b*u_2)*(w*v_1 + b*v_2) + abs(a)*u_1*v_1 def norm(u, w, a, b): r""" Returns the norm of the vector `u = (u_{1}, u_{2})` under the dot product defined by `u \cdot v = (wu_{1} + bu_{2})(w*v_{1} + bv_{2}) + |a|*u_{1}*v_{1}` where `u = (u_{1}, u_{2})` and `v = (v_{1}, v_{2})`. """ u_1, u_2 = u return sqrt(dot((u_1, u_2), (u_1, u_2), w, a, b))
[docs]def holzer(x, y, z, a, b, c): r""" Simplify the solution `(x, y, z)` of the equation `ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z^2 \geq \mid ab \mid` to a new reduced solution `(x', y', z')` such that `z'^2 \leq \mid ab \mid`. The algorithm is an interpretation of Mordell's reduction as described on page 8 of Cremona and Rusin's paper [1]_ and the work of Mordell in reference [2]_. References ========== .. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0. .. [2] Diophantine Equations, L. J. Mordell, page 48. """ if _odd(c): k = 2*c else: k = c//2 small = a*b*c step = 0 while True: t1, t2, t3 = a*x**2, b*y**2, c*z**2 # check that it's a solution if t1 + t2 != t3: if step == 0: raise ValueError('bad starting solution') break x_0, y_0, z_0 = x, y, z if max(t1, t2, t3) <= small: # Holzer condition break uv = u, v = base_solution_linear(k, y_0, -x_0) if None in uv: break p, q = -(a*u*x_0 + b*v*y_0), c*z_0 r = Rational(p, q) if _even(c): w = _nint_or_floor(p, q) assert abs(w - r) <= S.Half else: w = p//q # floor if _odd(a*u + b*v + c*w): w += 1 assert abs(w - r) <= S.One A = (a*u**2 + b*v**2 + c*w**2) B = (a*u*x_0 + b*v*y_0 + c*w*z_0) x = Rational(x_0*A - 2*u*B, k) y = Rational(y_0*A - 2*v*B, k) z = Rational(z_0*A - 2*w*B, k) assert all(i.is_Integer for i in (x, y, z)) step += 1 return tuple([int(i) for i in (x_0, y_0, z_0)])
[docs]def diop_general_pythagorean(eq, param=symbols("m", integer=True)): """ Solves the general pythagorean equation, `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. Returns a tuple which contains a parametrized solution to the equation, sorted in the same order as the input variables. Usage ===== ``diop_general_pythagorean(eq, param)``: where ``eq`` is a general pythagorean equation which is assumed to be zero and ``param`` is the base parameter used to construct other parameters by subscripting. Examples ======== >>> from sympy.solvers.diophantine import diop_general_pythagorean >>> from sympy.abc import a, b, c, d, e >>> diop_general_pythagorean(a**2 + b**2 + c**2 - d**2) (m1**2 + m2**2 - m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2) >>> diop_general_pythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2) (10*m1**2 + 10*m2**2 + 10*m3**2 - 10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4) """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "general_pythagorean": return _diop_general_pythagorean(var, coeff, param)
def _diop_general_pythagorean(var, coeff, t): if sign(coeff[var[0]**2]) + sign(coeff[var[1]**2]) + sign(coeff[var[2]**2]) < 0: for key in coeff.keys(): coeff[key] = -coeff[key] n = len(var) index = 0 for i, v in enumerate(var): if sign(coeff[v**2]) == -1: index = i m = symbols('%s1:%i' % (t, n), integer=True) ith = sum(m_i**2 for m_i in m) L = [ith - 2*m[n - 2]**2] L.extend([2*m[i]*m[n-2] for i in range(n - 2)]) sol = L[:index] + [ith] + L[index:] lcm = 1 for i, v in enumerate(var): if i == index or (index > 0 and i == 0) or (index == 0 and i == 1): lcm = ilcm(lcm, sqrt(abs(coeff[v**2]))) else: s = sqrt(coeff[v**2]) lcm = ilcm(lcm, s if _odd(s) else s//2) for i, v in enumerate(var): sol[i] = (lcm*sol[i]) / sqrt(abs(coeff[v**2])) return tuple(sol)
[docs]def diop_general_sum_of_squares(eq, limit=1): r""" Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Returns at most ``limit`` number of solutions. Usage ===== ``general_sum_of_squares(eq, limit)`` : Here ``eq`` is an expression which is assumed to be zero. Also, ``eq`` should be in the form, `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Details ======= When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be no solutions. Refer [1]_ for more details. Examples ======== >>> from sympy.solvers.diophantine import diop_general_sum_of_squares >>> from sympy.abc import a, b, c, d, e, f >>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345) {(15, 22, 22, 24, 24)} Reference ========= .. [1] Representing an integer as a sum of three squares, [online], Available: http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "general_sum_of_squares": return _diop_general_sum_of_squares(var, -coeff[1], limit)
def _diop_general_sum_of_squares(var, k, limit=1): # solves Eq(sum(i**2 for i in var), k) n = len(var) if n < 3: raise ValueError('n must be greater than 2') s = set() if k < 0 or limit < 1: return s sign = [-1 if x.is_nonpositive else 1 for x in var] negs = sign.count(-1) != 0 took = 0 for t in sum_of_squares(k, n, zeros=True): if negs: s.add(tuple([sign[i]*j for i, j in enumerate(t)])) else: s.add(t) took += 1 if took == limit: break return s
[docs]def diop_general_sum_of_even_powers(eq, limit=1): """ Solves the equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` where `e` is an even, integer power. Returns at most ``limit`` number of solutions. Usage ===== ``general_sum_of_even_powers(eq, limit)`` : Here ``eq`` is an expression which is assumed to be zero. Also, ``eq`` should be in the form, `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`. Examples ======== >>> from sympy.solvers.diophantine import diop_general_sum_of_even_powers >>> from sympy.abc import a, b >>> diop_general_sum_of_even_powers(a**4 + b**4 - (2**4 + 3**4)) {(2, 3)} See Also ======== power_representation() """ var, coeff, diop_type = classify_diop(eq, _dict=False) if diop_type == "general_sum_of_even_powers": for k in coeff.keys(): if k.is_Pow and coeff[k]: p = k.exp return _diop_general_sum_of_even_powers(var, p, -coeff[1], limit)
def _diop_general_sum_of_even_powers(var, p, n, limit=1): # solves Eq(sum(i**2 for i in var), n) k = len(var) s = set() if n < 0 or limit < 1: return s sign = [-1 if x.is_nonpositive else 1 for x in var] negs = sign.count(-1) != 0 took = 0 for t in power_representation(n, p, k): if negs: s.add(tuple([sign[i]*j for i, j in enumerate(t)])) else: s.add(t) took += 1 if took == limit: break return s ## Functions below this comment can be more suitably grouped under ## an Additive number theory module rather than the Diophantine ## equation module.
[docs]def partition(n, k=None, zeros=False): """ Returns a generator that can be used to generate partitions of an integer `n`. A partition of `n` is a set of positive integers which add up to `n`. For example, partitions of 3 are 3, 1 + 2, 1 + 1 + 1. A partition is returned as a tuple. If ``k`` equals None, then all possible partitions are returned irrespective of their size, otherwise only the partitions of size ``k`` are returned. If the ``zero`` parameter is set to True then a suitable number of zeros are added at the end of every partition of size less than ``k``. ``zero`` parameter is considered only if ``k`` is not None. When the partitions are over, the last `next()` call throws the ``StopIteration`` exception, so this function should always be used inside a try - except block. Details ======= ``partition(n, k)``: Here ``n`` is a positive integer and ``k`` is the size of the partition which is also positive integer. Examples ======== >>> from sympy.solvers.diophantine import partition >>> f = partition(5) >>> next(f) (1, 1, 1, 1, 1) >>> next(f) (1, 1, 1, 2) >>> g = partition(5, 3) >>> next(g) (1, 1, 3) >>> next(g) (1, 2, 2) >>> g = partition(5, 3, zeros=True) >>> next(g) (0, 0, 5) """ from sympy.utilities.iterables import ordered_partitions if not zeros or k is None: for i in ordered_partitions(n, k): yield tuple(i) else: for m in range(1, k + 1): for i in ordered_partitions(n, m): i = tuple(i) yield (0,)*(k - len(i)) + i
[docs]def prime_as_sum_of_two_squares(p): """ Represent a prime `p` as a unique sum of two squares; this can only be done if the prime is congruent to 1 mod 4. Examples ======== >>> from sympy.solvers.diophantine import prime_as_sum_of_two_squares >>> prime_as_sum_of_two_squares(7) # can't be done >>> prime_as_sum_of_two_squares(5) (1, 2) Reference ========= .. [1] Representing a number as a sum of four squares, [online], Available: http://schorn.ch/lagrange.html See Also ======== sum_of_squares() """ if not p % 4 == 1: return if p % 8 == 5: b = 2 else: b = 3 while pow(b, (p - 1) // 2, p) == 1: b = nextprime(b) b = pow(b, (p - 1) // 4, p) a = p while b**2 > p: a, b = b, a % b return (int(a % b), int(b)) # convert from long
[docs]def sum_of_three_squares(n): r""" Returns a 3-tuple `(a, b, c)` such that `a^2 + b^2 + c^2 = n` and `a, b, c \geq 0`. Returns None if `n = 4^a(8m + 7)` for some `a, m \in Z`. See [1]_ for more details. Usage ===== ``sum_of_three_squares(n)``: Here ``n`` is a non-negative integer. Examples ======== >>> from sympy.solvers.diophantine import sum_of_three_squares >>> sum_of_three_squares(44542) (18, 37, 207) References ========== .. [1] Representing a number as a sum of three squares, [online], Available: http://schorn.ch/lagrange.html See Also ======== sum_of_squares() """ special = {1:(1, 0, 0), 2:(1, 1, 0), 3:(1, 1, 1), 10: (1, 3, 0), 34: (3, 3, 4), 58:(3, 7, 0), 85:(6, 7, 0), 130:(3, 11, 0), 214:(3, 6, 13), 226:(8, 9, 9), 370:(8, 9, 15), 526:(6, 7, 21), 706:(15, 15, 16), 730:(1, 27, 0), 1414:(6, 17, 33), 1906:(13, 21, 36), 2986: (21, 32, 39), 9634: (56, 57, 57)} v = 0 if n == 0: return (0, 0, 0) v = multiplicity(4, n) n //= 4**v if n % 8 == 7: return if n in special.keys(): x, y, z = special[n] return _sorted_tuple(2**v*x, 2**v*y, 2**v*z) s, _exact = integer_nthroot(n, 2) if _exact: return (2**v*s, 0, 0) x = None if n % 8 == 3: s = s if _odd(s) else s - 1 for x in range(s, -1, -2): N = (n - x**2) // 2 if isprime(N): y, z = prime_as_sum_of_two_squares(N) return _sorted_tuple(2**v*x, 2**v*(y + z), 2**v*abs(y - z)) return if n % 8 == 2 or n % 8 == 6: s = s if _odd(s) else s - 1 else: s = s - 1 if _odd(s) else s for x in range(s, -1, -2): N = n - x**2 if isprime(N): y, z = prime_as_sum_of_two_squares(N) return _sorted_tuple(2**v*x, 2**v*y, 2**v*z)
[docs]def sum_of_four_squares(n): r""" Returns a 4-tuple `(a, b, c, d)` such that `a^2 + b^2 + c^2 + d^2 = n`. Here `a, b, c, d \geq 0`. Usage ===== ``sum_of_four_squares(n)``: Here ``n`` is a non-negative integer. Examples ======== >>> from sympy.solvers.diophantine import sum_of_four_squares >>> sum_of_four_squares(3456) (8, 8, 32, 48) >>> sum_of_four_squares(1294585930293) (0, 1234, 2161, 1137796) References ========== .. [1] Representing a number as a sum of four squares, [online], Available: http://schorn.ch/lagrange.html See Also ======== sum_of_squares() """ if n == 0: return (0, 0, 0, 0) v = multiplicity(4, n) n //= 4**v if n % 8 == 7: d = 2 n = n - 4 elif n % 8 == 6 or n % 8 == 2: d = 1 n = n - 1 else: d = 0 x, y, z = sum_of_three_squares(n) return _sorted_tuple(2**v*d, 2**v*x, 2**v*y, 2**v*z)
def power_representation(n, p, k, zeros=False): """ Returns a generator for finding k-tuples of integers, `(n_{1}, n_{2}, . . . n_{k})`, such that `n = n_{1}^p + n_{2}^p + . . . n_{k}^p`. Usage ===== ``power_representation(n, p, k, zeros)``: Represent non-negative number ``n`` as a sum of ``k`` ``p``th powers. If ``zeros`` is true, then the solutions is allowed to contain zeros. Examples ======== >>> from sympy.solvers.diophantine import power_representation Represent 1729 as a sum of two cubes: >>> f = power_representation(1729, 3, 2) >>> next(f) (9, 10) >>> next(f) (1, 12) If the flag `zeros` is True, the solution may contain tuples with zeros; any such solutions will be generated after the solutions without zeros: >>> list(power_representation(125, 2, 3, zeros=True)) [(5, 6, 8), (3, 4, 10), (0, 5, 10), (0, 2, 11)] For even `p` the `permute_sign` function can be used to get all signed values: >>> from sympy.utilities.iterables import permute_signs >>> list(permute_signs((1, 12))) [(1, 12), (-1, 12), (1, -12), (-1, -12)] All possible signed permutations can also be obtained: >>> from sympy.utilities.iterables import signed_permutations >>> list(signed_permutations((1, 12))) [(1, 12), (-1, 12), (1, -12), (-1, -12), (12, 1), (-12, 1), (12, -1), (-12, -1)] """ n, p, k = [as_int(i) for i in (n, p, k)] if n < 0: if p % 2: for t in power_representation(-n, p, k, zeros): yield tuple(-i for i in t) return if p < 1 or k < 1: raise ValueError(filldedent(''' Expecting positive integers for `(p, k)`, but got `(%s, %s)`''' % (p, k))) if n == 0: if zeros: yield (0,)*k return if k == 1: if p == 1: yield (n,) else: be = perfect_power(n) if be: b, e = be d, r = divmod(e, p) if not r: yield (b**d,) return if p == 1: for t in partition(n, k, zeros=zeros): yield t return if p == 2: feasible = _can_do_sum_of_squares(n, k) if not feasible: return if not zeros and n > 33 and k >= 5 and k <= n and n - k in ( 13, 10, 7, 5, 4, 2, 1): '''Todd G. Will, "When Is n^2 a Sum of k Squares?", [online]. Available: https://www.maa.org/sites/default/files/Will-MMz-201037918.pdf''' return if feasible is 1: # it's prime and k == 2 yield prime_as_sum_of_two_squares(n) return if k == 2 and p > 2: be = perfect_power(n) if be and be[1] % p == 0: return # Fermat: a**n + b**n = c**n has no solution for n > 2 if n >= k: a = integer_nthroot(n - (k - 1), p)[0] for t in pow_rep_recursive(a, k, n, [], p): yield tuple(reversed(t)) if zeros: a = integer_nthroot(n, p)[0] for i in range(1, k): for t in pow_rep_recursive(a, i, n, [], p): yield tuple(reversed(t + (0,) * (k - i))) sum_of_powers = power_representation def pow_rep_recursive(n_i, k, n_remaining, terms, p): if k == 0 and n_remaining == 0: yield tuple(terms) else: if n_i >= 1 and k > 0: for t in pow_rep_recursive(n_i - 1, k, n_remaining, terms, p): yield t residual = n_remaining - pow(n_i, p) if residual >= 0: for t in pow_rep_recursive(n_i, k - 1, residual, terms + [n_i], p): yield t
[docs]def sum_of_squares(n, k, zeros=False): """Return a generator that yields the k-tuples of nonnegative values, the squares of which sum to n. If zeros is False (default) then the solution will not contain zeros. The nonnegative elements of a tuple are sorted. * If k == 1 and n is square, (n,) is returned. * If k == 2 then n can only be written as a sum of squares if every prime in the factorization of n that has the form 4*k + 3 has an even multiplicity. If n is prime then it can only be written as a sum of two squares if it is in the form 4*k + 1. * if k == 3 then n can be written as a sum of squares if it does not have the form 4**m*(8*k + 7). * all integers can be written as the sum of 4 squares. * if k > 4 then n can be partitioned and each partition can be written as a sum of 4 squares; if n is not evenly divisible by 4 then n can be written as a sum of squares only if the an additional partition can be written as sum of squares. For example, if k = 6 then n is partitioned into two parts, the first being written as a sum of 4 squares and the second being written as a sum of 2 squares -- which can only be done if the condition above for k = 2 can be met, so this will automatically reject certain partitions of n. Examples ======== >>> from sympy.solvers.diophantine import sum_of_squares >>> list(sum_of_squares(25, 2)) [(3, 4)] >>> list(sum_of_squares(25, 2, True)) [(3, 4), (0, 5)] >>> list(sum_of_squares(25, 4)) [(1, 2, 2, 4)] See Also ======== sympy.utilities.iterables.signed_permutations """ for t in power_representation(n, 2, k, zeros): yield t
def _can_do_sum_of_squares(n, k): """Return True if n can be written as the sum of k squares, False if it can't, or 1 if k == 2 and n is prime (in which case it *can* be written as a sum of two squares). A False is returned only if it can't be written as k-squares, even if 0s are allowed. """ if k < 1: return False if n < 0: return False if n == 0: return True if k == 1: return is_square(n) if k == 2: if n in (1, 2): return True if isprime(n): if n % 4 == 1: return 1 # signal that it was prime return False else: f = factorint(n) for p, m in f.items(): # we can proceed iff no prime factor in the form 4*k + 3 # has an odd multiplicity if (p % 4 == 3) and m % 2: return False return True if k == 3: if (n//4**multiplicity(4, n)) % 8 == 7: return False # every number can be written as a sum of 4 squares; for k > 4 partitions # can be 0 return True